Projectile range formula Steps:

• Horizontal velocity component: $V_x = V \cos \theta$
• Time of flight to return to same level: $t = \frac{2 V \sin \theta}{g}$
• Range: $R = V_x \times t = V \cos \theta \times \frac{2 V \sin \theta}{g}$
• Simplify: $R = \frac{V^2 \sin 2\theta}{g}$ (using $\sin$)

Why B is correct:

• Matches the derived standard formula for horizontal range in projectile motion under gravity.

Why the others are wrong:

• A: Lacks a factor of $V$ and 2, underestimates range.
• C: Uses $\sin \theta \cos \theta$ instead of $\sin 2\theta$, missing factor of 2.
• D: Results in $\frac{V^2 \tan \theta}{g}$, valid for inclined plane but not horizontal.

Final answer: B